Solve 1/x-1/10=1/x+5

$$\frac{ 1 }{ x } - \frac{ 1 }{ 10 } = \frac{ 1 }{ x+5 }$$

Solve for x

$$x=-10$$

$$x=5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x+5\right), the least common multiple of x,10,x+5.

$$10x+50+10x\left(x+5\right)\left(-\frac{1}{10}\right)=10x$$

Multiply 10 and -\frac{1}{10} to get -1.

$$10x+50-x\left(x+5\right)=10x$$

Use the distributive property to multiply -x by x+5.

$$10x+50-x^{2}-5x=10x$$

Combine 10x and -5x to get 5x.

$$5x+50-x^{2}=10x$$

Subtract 10x from both sides.

$$5x+50-x^{2}-10x=0$$

Combine 5x and -10x to get -5x.

$$-5x+50-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}-5x+50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-50=-50$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -50.

$$1,-50 2,-25 5,-10$$

Calculate the sum for each pair.

$$1-50=-49 2-25=-23 5-10=-5$$

The solution is the pair that gives sum -5.

$$a=5 b=-10$$

Rewrite -x^{2}-5x+50 as \left(-x^{2}+5x\right)+\left(-10x+50\right).

$$\left(-x^{2}+5x\right)+\left(-10x+50\right)$$

Factor out x in the first and 10 in the second group.

$$x\left(-x+5\right)+10\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(x+10\right)$$

To find equation solutions, solve -x+5=0 and x+10=0.

$$x=5 x=-10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x+5\right), the least common multiple of x,10,x+5.

$$10x+50+10x\left(x+5\right)\left(-\frac{1}{10}\right)=10x$$

Multiply 10 and -\frac{1}{10} to get -1.

$$10x+50-x\left(x+5\right)=10x$$

Use the distributive property to multiply -x by x+5.

$$10x+50-x^{2}-5x=10x$$

Combine 10x and -5x to get 5x.

$$5x+50-x^{2}=10x$$

Subtract 10x from both sides.

$$5x+50-x^{2}-10x=0$$

Combine 5x and -10x to get -5x.

$$-5x+50-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}-5x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -5 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-1\right)\times 50}}{2\left(-1\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-1\right)\times 50}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-5\right)±\sqrt{25+4\times 50}}{2\left(-1\right)}$$

Multiply 4 times 50.

$$x=\frac{-\left(-5\right)±\sqrt{25+200}}{2\left(-1\right)}$$

Add 25 to 200.

$$x=\frac{-\left(-5\right)±\sqrt{225}}{2\left(-1\right)}$$

Take the square root of 225.

$$x=\frac{-\left(-5\right)±15}{2\left(-1\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±15}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{5±15}{-2}$$

Now solve the equation x=\frac{5±15}{-2} when ± is plus. Add 5 to 15.

$$x=\frac{20}{-2}$$

Divide 20 by -2.

$$x=-10$$

Now solve the equation x=\frac{5±15}{-2} when ± is minus. Subtract 15 from 5.

$$x=-\frac{10}{-2}$$

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=-10 x=5$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by 10x\left(x+5\right), the least common multiple of x,10,x+5.

$$10x+50+10x\left(x+5\right)\left(-\frac{1}{10}\right)=10x$$

Multiply 10 and -\frac{1}{10} to get -1.

$$10x+50-x\left(x+5\right)=10x$$

Use the distributive property to multiply -x by x+5.

$$10x+50-x^{2}-5x=10x$$

Combine 10x and -5x to get 5x.

$$5x+50-x^{2}=10x$$

Subtract 10x from both sides.

$$5x+50-x^{2}-10x=0$$

Combine 5x and -10x to get -5x.

$$-5x+50-x^{2}=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$-5x-x^{2}=-50$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}-5x=-50$$

Divide both sides by -1.

$$\frac{-x^{2}-5x}{-1}=-\frac{50}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{5}{-1}\right)x=-\frac{50}{-1}$$

Divide -5 by -1.

$$x^{2}+5x=-\frac{50}{-1}$$

Divide -50 by -1.

$$x^{2}+5x=50$$

Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=50+\left(\frac{5}{2}\right)^{2}$$

Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+5x+\frac{25}{4}=50+\frac{25}{4}$$

Add 50 to \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{225}{4}$$

Factor x^{2}+5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{225}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{225}{4}}$$

Simplify.

$$x+\frac{5}{2}=\frac{15}{2} x+\frac{5}{2}=-\frac{15}{2}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=5 x=-10$$